Create model based api without any extra code.
Project description
# django-easy-api
Now there is no need to create apis for common CRUD operation for any model. By using django-easy-api its really easy. Kindly follow the below steps for including this library into your project.
Create Your django project.
Add easy_api in settings.py
INSTALLED_APPS = [ . . . ‘easy_api’ ]
Create your django app
Now in models.py follow create your model by inheriting EasyAPI model from easy_api app.
from django.db import models from easy_api.models import EasyAPI
- class MyModel(EasyAPI):
# Your Requried field here.
Migrate your models.
This is the last step. You need to add this model in your app’s urls.py
from django.urls import path from .models import MyModel
- urlpatterns = [
path(‘myurl/’, MyModel.as_view()),
]
7. That’s it. Now you will have common GET/POST/PUT/DELETE methods on your model. Also you can attach query_params only for id in your url. For example:
8. You can also override get, post, put and delete method according to your need. For example:
from django.db import models from easy_api.models import EasyAPI
- class MyModel(EasyAPI):
# Your Requried field here.
- def get(self, request):
#Your logic
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